Which of the following best represents the temperature versus heat supplied graph for water, in the range of -20^circmathrmC$-20^{\circ}\mathrm{C}$ to 120^circmathrmC$120^{\circ}\mathrm{C}$ ?

### Related Formula P = fracdmdt S Delta T$$P = \frac{dm}{dt} S \Delta T$$ P_textheater = textRate of combustion times textHeat of combustion$$P_{\text{heater}} = \text{Rate of combustion} \times \text{Heat of combustion}$$ ### Core Logic Water flow rate = 5text l/min = frac560text kg/s = frac112text kg/s$5\text{ l/min} = \frac{5}{60}\text{ kg/s} = \frac{1}{12}\text{ kg/s}$. The power required to heat the water is: P_textheater = left(fracdmdtright)_textwater cdot S cdot Delta T$$P_{\text{heater}} = \left(\frac{dm}{dt}\right)_{\text{water}} \cdot S \cdot \Delta T$$ P_textheater = frac112 times 4200 times (87 - 27) = frac112 times 4200 times 60text W$$P_{\text{heater}} = \frac{1}{12} \times 4200 \times (87 - 27) = \frac{1}{12} \times 4200 \times 60\text{ W}$$ ### Step 1: Solving for Gas Consumption Let the rate of consumption of gas be xtext g/s$x\text{ g/s}$. Heat generated by the gas per second must equal the power required to heat the water: x times 5.0 times 10^4 = frac112 times 4200 times 60$$x \times 5.0 \times 10^{4} = \frac{1}{12} \times 4200 \times 60$$ x times 5.0 times 10^4 = 4200 times 5$$x \times 5.0 \times 10^{4} = 4200 \times 5$$ x = frac4200 times 55 times 10^4 = 4200 times 10^-4$$x = \frac{4200 \times 5}{5 \times 10^{4}} = 4200 \times 10^{-4}$$ x = 0.42text g/s$$x = 0.42\text{ g/s}$$

### Pattern Recognition Equate the thermal power absorbed by fluid (dm/dt cdot c cdot Delta T$dm/dt \cdot c \cdot \Delta T$) with the power output of fuel (dm_textfuel/dt cdot L_textcombustion$dm_{\text{fuel}}/dt \cdot L_{\text{combustion}}$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermal Properties of Matter

### Related Formula Delta l = l_0 alpha Delta T$$\Delta l = l_0 \alpha \Delta T$$ l_textfinal = l_1 + l_2 + Delta l_1 + Delta l_2$$l_{\text{final}} = l_1 + l_2 + \Delta l_1 + \Delta l_2$$ ### Core Logic Let the original lengths be l_0A$l_{0A}$ (aluminium) and l_0S$l_{0S}$ (steel). They have same length, so l_0A = l_0S = 60text cm$l_{0A} = l_{0S} = 60\text{ cm}$. Change in temperature Delta T = 100^circtextC - 30^circtextC = 70^circtextC$\Delta T = 100^{\circ}\text{C} - 30^{\circ}\text{C} = 70^{\circ}\text{C}$. Expansion of Aluminium: Delta l_A = l_0 alpha_A Delta T$\Delta l_A = l_0 \alpha_A \Delta T$ Expansion of Steel: Delta l_S = l_0 alpha_S Delta T$\Delta l_S = l_0 \alpha_S \Delta T$ ### Step 1: Calculate Final Length l_textfinal = l_0(1 + alpha_A Delta T) + l_0(1 + alpha_S Delta T)$$l_{\text{final}} = l_0(1 + \alpha_A \Delta T) + l_0(1 + \alpha_S \Delta T)$$ l_textfinal = l_0 [2 + (alpha_A + alpha_S)Delta T]$$l_{\text{final}} = l_0 [2 + (\alpha_A + \alpha_S)\Delta T]$$ l_textfinal = 60 left[ 2 + (24 times 10^-6 + 12 times 10^-6) times 70 right]$$l_{\text{final}} = 60 \left[ 2 + (24 \times 10^{-6} + 12 \times 10^{-6}) \times 70 \right]$$ l_textfinal = 60 left[ 2 + (36 times 10^-6) times 70 right]$$l_{\text{final}} = 60 \left[ 2 + (36 \times 10^{-6}) \times 70 \right]$$ l_textfinal = 60 [2 + 0.00252] = 120 + 60(0.00252) = 120 + 0.1512 = 120.15text cm$$l_{\text{final}} = 60 [2 + 0.00252] = 120 + 60(0.00252) = 120 + 0.1512 = 120.15\text{ cm}$$

### Pattern Recognition For composite rods connected end-to-end, expansions simply add up. Convert 1.2 times 10^-5$1.2 \times 10^{-5}$ to 12 times 10^-6$12 \times 10^{-6}$ immediately for easy mental addition. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermal Properties of Matter

### Related Formula H = fracDelta TR_texteq, quad R = fraclKA$$H = \frac{\Delta T}{R_{\text{eq}}}, \quad R = \frac{l}{KA}$$ ### Core Logic

Let thermal resistance of rod y be 3R$3R$ and rod x be R$R$ (since K_x = 3K_y$K_x = 3K_y$). Using thermal current equations: H = frac100 - 40frac11R2$$H = \frac{100 - 40}{\frac{11R}{2}}$$ H = frac100 - T_BR, quad H = fracT_E - 403R$$H = \frac{100 - T_B}{R}, \quad H = \frac{T_E - 40}{3R}$$ Solving yields T_B = 89^circtextC$T_B = 89^{circ}\text{C}$ and T_E = 73^circtextC$T_E = 73^{circ}\text{C}$. ### Pattern Recognition Sees: Composite thermal conduction network with series/parallel segments. Shortcut: Calculate equivalent thermal resistances and equate heat currents across junctions. Check: Matches option (1). ✓ ### Chapter Mix Class 11 Physics: Thermal Properties of Matter

### Related Formula Q = mcDelta T$Q = mc\Delta T$ Q = mL$Q = mL$ ### Core Logic Heat gained by ice to reach final temperature T$T$ must equal the heat lost by the water to reach temperature T$T$. ### Step 1: Energy Balance Equation Let T$T$ be the final temperature of the mixture. Heat gained by ice to reach 0^circmathrmC$0^{\circ}\mathrm{C}$ + Heat to melt + Heat to reach T$T$ = Heat lost by water. 10 times 3.36 times 10^5 + 10 times 2100 times 10 + 10 times 4200 times (T - 0) = 100 times 4200 times (25 - T)$$10 \times 3.36 \times 10^{5} + 10 \times 2100 \times 10 + 10 \times 4200 \times (T - 0) = 100 \times 4200 \times (25 - T)$$ ### Step 2: Solving for Final Temperature 3.36 times 10^6 + 2.1 times 10^5 + 4.2 times 10^4 T = 1.05 times 10^7 - 4.2 times 10^5 T$$3.36 \times 10^{6} + 2.1 \times 10^{5} + 4.2 \times 10^{4} T = 1.05 \times 10^{7} - 4.2 \times 10^{5} T$$ 3.57 times 10^6 + 4.2 times 10^4 T = 1.05 times 10^7 - 4.2 times 10^5 T$$3.57 \times 10^{6} + 4.2 \times 10^{4} T = 1.05 \times 10^{7} - 4.2 \times 10^{5} T$$ Solving for T$T$: 4.62 times 10^5 T = 6.93 times 10^6$$4.62 \times 10^{5} T = 6.93 \times 10^{6}$$ T = 15^circmathrmC$$T = 15^{\circ}\mathrm{C}$$ ### Step 3: Calculating Decrement The decrement in the temperature of water is: Delta T = 25 - 15 = 10^circmathrmC$$\Delta T = 25 - 15 = 10^{\circ}\mathrm{C}$$ ### Pattern Recognition When mixing ice and water, track the 3 stages for ice (warm to 0, melt, warm to T). Equate total heat gained to heat lost by water. Always solve for T first, then check what the question actually asks (decrement). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermal Properties of Matter